Easy math question, mind went blank,

[speg]

How do you solve an equation with a root in it?!

For example:

2x+3x^(1/2) = 20


My mind has gone blank, can someone remind me how to solve this?

 

[gopunk]

square both sides and solve the quadratic?

 

[Dark4ng3l]

2x+3x^(1/2) = 20
3x^(1/2) = 20-2x

square both sides and solve with the quadratic formula.

 

[speg]

If I do that, [2x+3x^(1/2)]^2 expands out and has a 12x(1/2) in it, Isn't that supposed to be gone? Ugh, Im dumb 🙁

EDIT:
Ok, order has been restored to my mind. Sorry about that 😱

 

[Orsorum]

Originally posted by: gopunk
square both sides and solve the quadratic?

That's what I thought initially, but you'd still be stuck with a...no, that could work. Subtract 20 from both sides, and treat it as the quadratic 2(x^1/2)^2 + 3x^(1/2)-20 = 0; when you find the solutions for x^1/2, just take the roots.

Don't know if that would work, you could end up with some funky imaginary numbers.

 

[NaughtyusMaximus]

just use the quadratic formula on it, treating the x ^1 as the largest power term (normally x^2 in the quadratic forumula), and the x^(1/2) as as the lower power term (normally x^1).

 

[newt]

Originally posted by: Dark4ng3l
2x+3x^(1/2) = 20
3x^(1/2) = 20-2x

square both sides and solve with the quadratic formula.

right, x=14.694933

 

[Legendary]

(20-2x)/3 = x^(1/2)
9x = 400-80x+4x^2
4x^2 - 89x + 400 = 0
um...wow.
x = -b +- sqrt(b^2-4ac) / 2a

Enjoy. 😀

 

[Analog]

Originally posted by: Dark4ng3l
2x+3x^(1/2) = 20
3x^(1/2) = 20-2x

square both sides and solve with the quadratic formula.

yep